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Integration Question

  1. Sep 2, 2010 #1

    When we use the integration by parts to find the integral of xcosx dx, we assume that cosx dx = du and integrate both sides to find u
    When we integrate
    du = cosxdx
    we find that u = sinx
    the question is why don't we take the integration constant C into account?
  2. jcsd
  3. Sep 2, 2010 #2
    You can take it into account if you want, you'll get the same result (except for a final integration constant of course), so it's usually better to assume this constant zero in the intermediate steps.
  4. Sep 2, 2010 #3
    But it may or mayn't be zero, therefore the final integration is different if it's not zero.
  5. Sep 2, 2010 #4
    I indeed am in need of your help
  6. Sep 2, 2010 #5


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    Science Advisor

    The point is that it cancels out. Without the integration constant:
    [tex]\int{udv} = uv-\int{vdu}[/tex]
    With the integration constant:
    [tex]\intudv = u(v+C)-\int{(v+C)du} = uv+uC-\int{vdu}-C\int{du} = uv +uC-\int{vdu}-uC = uv-\int{vdu}[/tex]
  7. Sep 2, 2010 #6
    Wow, that's real helpful, thank you
  8. Sep 20, 2010 #7
    For this integrand it is much better to let u = x and dv = Cos x
    Your choice will work too, but if x is raised to a power the clear choice is
    u = x^n and dv = Cos x
    This integral is also best done with Tabular Integration By Parts
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